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2 years ago

A hazard sign has 3 identical

1 answer:

5 0

Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.

Perimeter is the sum of all the sides of a figure.

For a parallelogram:

P = 2*length + 2*width

So, we need to determine width and length of the stripe.

Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is

$h=\sqrt{18^{2}+6^{2}}$

$h=\sqrt{360}$

h = 19 in

Perimeter of the first stripe is

P = (2*19) + (2*3)

P = 44 inches

The hazard sign has 3 stripes. So total perimeter is

$P_{t}=$ 44 + 44 + 44

$P_{t}=$ 132 inches

To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.

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If revenue is $3,000 and operating expenses are $4,000, cash flow equals _____. -$1,000 $1,000 -$7,000 $7,000

valkas [14]

Answer:

-$1,000

Explanation:

$3000 - $4000

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3 years ago

You find a micrometer (a tool used to

MakcuM [25]

Explanation:

A micrometer is a measuring device or an instrument which is used to measure very minute measurements very accurately and precisely. It is mathematical tool used to provide accurate measurement for any mechanical components.

Now, if the micrometer that I have found is badly bent, it would provide faulty or wrong measurements both in terms of precision and accuracy when compared to a high quality meter stick.

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3 years ago

A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta

Blababa [14]

Explanation:

Given that,

The slope of the ramp, $\theta=8^{\circ}$

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

$W=\dfrac{1}{2}m(v^2-u^2)=0$

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

$W=Fd\ cos\theta$

Here, $\theta=90+8=98^{\circ}$

$W=mgd\ cos\theta$

$W=60\times 9.8\times 300\ cos(98)$

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

$W_f=-W$

$W_f=24550.13\ J$

Hence, this is the required solution.

6 0

3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

If your front lawn is 18.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snow flakes every min

Ivanshal [37]

<span>First, we need to determine the entire area of your front line by multiplying its length times its width.
18.0*20.0 = 360.0 square feet
We can use the rate of accumulation of snow, combined with this figure, to determine how much snow accumulates on your lawn per minute.
360.0 sq ft * 1050 flakes/min/sq ft = 378,000 flakes/min
We can then use the mass of a snowflake to calculate total snow accumulation per minute.
378,000 flakes/min * 2.00 mg/flake = 756,000 mg/min
Finally, we can use this number to determine accumulation per hour.
756,000 mg/min * 60 min/hr =
45,360,000 mg/hr</span>

8 0

2 years ago